course Phy 121

001. typewriter notationNote that there are six questions in this exercise. Be sure to continue scrolling down until you get to the end of the exercise.

*********************************************

Question: `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4). Then evaluate each expression for x = 2.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

X – 2 / x + 4 =

X – 2 / 2 + 4 =

X – 1 + 4 =

2 – 1 + 4 =

1 + 4 = 5

(x - 2) / (x + 4) =

(2 – 2) / (2 + 4) =

(0) / (6) = 0

confidence rating #$&* 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`aThe order of operations dictates that grouped expressions must be evaluated first, that exponentiation must be done before multiplication or division, which must be done before addition or subtraction.

It makes a big difference whether you subtract the 2 from the x or divide the -2 by 4 first. If there are no parentheses you have to divide before you subtract. Substituting 2 for x we get

2 - 2 / 2 + 4

= 2 - 1 + 4 (do multiplications and divisions before additions and subtractions)

= 5 (add and subtract in indicated order)

If there are parentheses you evaluate the grouped expressions first:

(x - 2) / (x + 4) = (2 - 2) / ( 2 + 4 ) = 0 / 6 = 0.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

*********************************************

Question: `q002. Explain the difference between 2 ^ x + 4 and 2 ^ (x + 4). Then evaluate each expression for x = 2.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

2^x + 4 =

2^2 + 4 =

4 + 4 = 8

2^ (x + 4) =

2^ (2 + 4) =

2^6 = 64

confidence rating #$&* 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a2 ^ x + 4 indicates that you are to raise 2 to the x power before adding the 4.

2 ^ (x + 4) indicates that you are to first evaluate x + 4, then raise 2 to this power.

If x = 2, then

2 ^ x + 4 = 2 ^ 2 + 4 = 2 * 2 + 4 = 4 + 4 = 8.

and

2 ^ (x + 4) = 2 ^ (2 + 4) = 2 ^ 6 = 2*2*2*2*2*2 = 64.

*********************************************

Question: `q003. What is the numerator of the fraction in the expression x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x? What is the denominator? What do you get when you evaluate the expression for x = 2?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Numerator =

x – 3 =

2 – 3 = -1

Denominator =

[(2x-5)^2 *3x +1] -2 + 7x=

[(2(2) -5)^2 *3(2) + 1] -2 + 7(2)=

[(4-5)^2 * 6 + 1] -2 + 14=

[(-1)^2 * 7] -16=

[1*7] -16=

7-16= -9

confidence rating #$&* 1 – clearly worked that one wrong (didn’t see that the 3 was divided by anything). Worked it out again on paper….. got it right that time

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`aThe numerator is 3. x isn't part of the fraction. / indicates division, which must always precede subtraction. Only the 3 is divided by [ (2x-5)^2 * 3x + 1 ] and only [ (2x-5)^2 * 3x + 1 ] divides 3.

If we mean (x - 3) / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x we have to write it that way.

The preceding comments show that the denominator is [ (2x-5)^2 * 3x + 1 ]

Evaluating the expression for x = 2:

- 3 / [ (2 * 2 - 5)^2 * 3(2) + 1 ] - 2 + 7*2 =

2 - 3 / [ (4 - 5)^2 * 6 + 1 ] - 2 + 14 = evaluate in parenthese; do multiplications outside parentheses

2 - 3 / [ (-1)^2 * 6 + 1 ] -2 + 14 = add inside parentheses

2 - 3 / [ 1 * 6 + 1 ] - 2 + 14 = exponentiate in bracketed term;

2 - 3 / 7 - 2 + 14 = evaluate in brackets

13 4/7 or 95/7 or about 13.57 add and subtract in order.

The details of the calculation 2 - 3 / 7 - 2 + 14:

Since multiplication precedes addition or subtraction the 3/7 must be done first, making 3/7 a fraction. Changing the order of the terms we have

2 - 2 + 14 - 3 / 7 = 14 - 3/7 = 98/7 - 3/7 = 95/7.

COMMON STUDENT QUESTION: ok, I dont understand why x isnt part of the fraction? And I dont understand why only the brackets are divided by 3..why not the rest of the equation?

INSTRUCTOR RESPONSE: Different situations give us different algebraic expressions; the situation dictates the form of the expression.

If the above expression was was written otherwise it would be a completely different expression and most likely give you a different result when you substitute.

If we intended the numerator to be x - 3 then the expression would be written (x - 3) / [(2x-5)^2 * 3x + 1 ] - 2 + 7x, with the x - 3 grouped.

If we intended the numerator to be the entire expression after the / the expression would be written x - 3 / [(2x-5)^2 * 3x + 1 - 2 + 7x ].

STUDENT COMMENT: I wasn't sure if the numerator would be 3 or -3. or is the subtraction sign just that a sign in this case?

INSTRUCTOR RESPONSE: In this case you would regard the - sign as an operation to be performed between the value of x and the value of the fraction, rather than as part of the numerator. That is, you would regard x - 3 / [ (2x-5)^2 * 3x + 1 ] as a subtraction of the fraction 3 / [ (2x-5)^2 * 3x + 1 ] from the term x.

STUDENT QUESTION: There was another question I had about this problem that wasn’t addressed. At the end when you changed the order of operation from

2 – 2 + 14 – 3/7 = 14 – 3/7

where did the 98/7 – 3/7 come into play before the end solution of 95/7? I must have forgotten how to do this part.

INSTRUCTOR RESPONSE: It's not clear how you can get 95/7 without this step.

To do the subtraction 14 - 3/7 both terms must be expressed in terms of a common denominator. The most convenient common denominator is 7.

So 14 must be expressed with denominator 7. This is accomplished by multiplying 14 by 7 / 7, obtaining 14 * 7 / 7 = 98 / 7. Since 7/7 = 1, we have just multiplied 14 by 1. We chose to use 7 / 7 in order to give us the desired denominator 7.

Thus our subtraction is

14 - 3/7 =

98/7 - 3/7 =

(98 - 3) / 7 =

95 /7.

STUDENT COMMENT

It took me a while to think thru this one especially when I got to working with the fraction. Fractions have always been my

weak spot. Any tips to make working with fractions a little easier is greatly appreciated.

INSTRUCTOR RESPONSE

Fractions are seriously undertaught in our schools, so your comment is not unusual.

I have to focus my attention on the subject matter of my courses, and while I do address it to a point, I don't have time to do justice to the subject of fractions. In any case , to do so would be redundant on my part, since there are a lot of excellent resources on the Internet.

I suggest you search the Web using something like 'review of fractions', and find something appropriate to your needs. You should definitely review the topic, as should 95% of all students entering your course.

STUDENT COMMENT

I think I am confused on why the Numerator is not the top portion and denominator the bottom portion of the problem.

INSTRUCTOR RESPONSE

Everything is on one line so there is no top or bottom in the given expression. A numerator and denominator are determined by a division of two expressions.

As we know, a denominator divides a numerator. In the given expression the division sign occurs between the 3 and the [ (2x-5)^2 * 3x + 1 ], so 3 is the numerator and [ (2x-5)^2 * 3x + 1 ] is the denominator.

x is not divided by the denominator, since the division occurs before the subtraction. For the same reason the -2 + 7x is not involved in the division. So neither the x nor the -2 + 7 x is part of the fractional expression.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

See above comments

comments should be copied here; however they were noted and OK

------------------------------------------------

Self-critique rating #$&*

*********************************************

Question: `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(X-5) is raised to the 2

(x-5)^2=

(4-5)^2=

-1^2 = 1

Then multiply by x

1*4 = 4

Then subtract 1

4-1 = 3

Then add 3/x

3 + ¾ =

3 ¾

Then subtract 2

3 ¾ - 2 =

1 ¾

confidence rating #$&* 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`aWe get

(4-5)^2 * 4 - 1 + 3 / 4 - 2

= (-1)^2 * 4 - 1 + 3 / 4 - 2 evaluating the term in parentheses

= 1 * 4 - 1 + 3 / 4 - 2 exponentiating (2 is the exponent, which is applied to -1 rather than multiplying the 2 by 4

= 4 - 1 + 3/4 - 2 noting that 3/4 is a fraction and adding and subtracting in order we get

= 1 3/4 = 7 /4 (Note that we could group the expression as 4 - 1 - 2 + 3/4 = 1 + 3/4 = 1 3/4 = 7/4).

COMMON ERROR:

(4 - 5) ^ 2*4 - 1 + 3 / 4 - 2 =

-1 ^ 2*4 - 1 + 3 / 4-2 =

-1 ^ 8 -1 + 3 / 4 - 2.

INSTRUCTOR COMMENTS:

There are two errors here. In the second step you can't multiply 2 * 4 because you have (-1)^2, which must be done first. Exponentiation precedes multiplication.

Also it isn't quite correct to write -1^2*4 at the beginning of the second step. If you were supposed to multiply 2 * 4 the expression would be (-1)^(2 * 4).

Note also that the -1 needs to be grouped because the entire expression (-1) is taken to the power. -1^8 would be -1 because you would raise 1 to the power 8 before applying the - sign, which is effectively a multiplication by -1.

STUDENT QUESTION: if it's read (-1)^8 it would be 1 or would you apply the sign afterward even if it is grouped and it be a -1?

INSTRUCTOR RESPONSE: The 8th power won't occur in this problem, of course, but you ask a good question.

-1^8 would require raising 1 to the 8th power, then applying the negative sign, and the result would be -1.

(-1)^8 would be the 8th power of -1, which as you see would be 1.

STUDENT COMMENT: I think it would be easier to visualize what your trying to raise to the exponent if you actually put parenthesis around the 2, that part seems to get tricky on the computer.

INSTRUCTOR RESPONSE: The expression was intentionally written to be misleading and make the point that, to avoid ambiguity, order of operations apply strictly, no matter what the expression looks like.

Normally, for clarity, the parentheses would be included. They aren't necessary, but when helpful it's a good idea to include them. You can, of course, have too many parentheses in an expression, making it harder than necessary to sort out. In practice we try to strike a balance.

The original expression was

(x - 5) ^ 2x-1 + 3 / x-2

White spaces make no difference in how an expression is evaluated, but they can help show the structure; e.g.,

(x - 5)^2 * x - 1 + 3 / x -2

is a visual improvement over the original. The * between the 2 and the x is not strictly necessary, but is also helpful.

((((x - 5) ^ 2)) * x) - 1 + (3 / x) - 2

verges on having too many parentheses at the beginning; it does help clarify the 3 / x.

STUDENT COMMENT

Although I read through your explanation and do see the point you are making, that 2x is actually 2 * x, I still think that

(-1) should be raised to 2x rather than 2. Kaking the answer -11/4, not 7/4.

INSTRUCTOR RESPONSE

When the expression (x - 5) ^ 2x-1 + 3 / x-2 is copied and pasted into a computer algebra system it is translated as

This notation is universal and unambiguous. Any deviation from strict interpretation (which does occur among some authors and among manufacturers of some calculators) tends to result in ambiguity and confusion.

STUDENT COMMENT

While I do understand what you are trying to relate, I will continue to make these mistakes on more than one occasion and will not penalize myself for not rewriting years of mathematics because of a syntax issue in an online class.

INSTRUCTOR RESPONSE

I don't penalize errors in typed notation when the intent is clear (though I will sometimes point out these errors), and when you take your tests you'll be writing them out by hand and this won't be an issue.

However this is not a syntax issue in an online class. This is the order of operations, as it has been since algebra was developed hundreds of years ago, and it's completely consistent with the mathematics you appear to know (quite well).

As stated here, if you use the wrong syntax in any computer algebra system, your expression will not be interpreted correctly. For this reason alone you need to understand the notation.

For this and other valid reasons you need to understand how the order of operations are represented in 'linear' fashion (i.e., 'typewriter notation') and to correctly interpret expressions written in this notatation.

Any mathematics that has been learned correctly is completely consistent with the order of operations and with the notation used in this course. If the mathematics you've learned was inconsistent with the order of operations (and I don't believe this is so in your case, but it is with many students), then you would need to adjust your thinking. Fortunately this is very easy to do. Interpret expressions literally, assume nothing, and everything works out.

You will also find that the notation quickly becomes easy to read and use, and that it expands your comprehension of all mathematical notation.

STUDENT COMMENT

I used -1^(2*4). I didn't realize that was doing multiplication before exponents. All of this typewriter notations seems ambiguous to me but I think that had I seen the expression in standard notation I would probably have made the same mistake in this instance. If I were writing this expression I would probably use a parenthesis or * to show the necessary separation.

INSTRUCTOR RESPONSE

Parentheses, even when they aren't strictly necessary, are often useful to clarify the expression. An parentheses, even when not necessary, are part of the order of operations.

Spacing is not part of the order of operations. An expression has the same meaning even if all spaces are removed.

However as long as an expression is correctly formed, spacing as well as parentheses can certainly be used to make it more readable.

I don't go to any trouble in this exercise to make the expressions readable, since my goal here is to make the point about order of operations, which give an expression its unambiguous meaning.

However in most of the documents you will be working with, I do make an effort to clarify the meanings of expressions through their formatting, often using unnecessary parentheses and spacing to help clarify meaning.

Certainly I encourage you to do the same.

STUDENT QUESTION

I didn’t separate the ¾ as a stand alone fraction, I am confused about why you don’t treat it as an equation that the

denominator isn’t treated as a denominator.

INSTRUCTOR RESPONSE

Your work was good throughout most of this problem. You did forget to copy down a -1 in one of the early steps, but otherwise followed the order of operations correctly until nearly the last step.

However near the end you said that 4+3/4-2=7/2.

You appear to have performed the addition 4 + 3 and the subtraction 4 - 2 before dividing. However the division has to be done first.

The division sign is between the 3 and the 4, so the division is 3/4, and that gives you the fraction 3/4.

Therefore the expression 4+3/4-2 tells you to 'add 3/4 to 4 then subtract 2'.

When actually writing this out we would probably include parentheses. That wasn't done here, as it would have defeated the point being made about order of operations, but for clarity we might have written

4 + (3/4) - 2.

The parentheses are not necessary around the 3/4, since the order of operations is sufficient to unambiguously define the result, but they do make the expression easier to read and reduce the likelihood of error.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): took my time with this one…. Worked out a little better

------------------------------------------------

Self-critique rating #$&*

*********************************************

Question: `q005. At the link

http://www.vhcc.edu/dsmith/genInfo/introductory problems/typewriter_notation_examples_with_links.htm

(copy this path into the Address box of your Internet browser; alternatively use the path

http://vhmthphy.vhcc.edu/ > General Information > Startup and Orientation (either scroll to bottom of page or click on Links to Supplemental Sites) > typewriter notation examples

and you will find a page containing a number of additional exercises and/or examples of typewriter notation.Locate this site, click on a few of the links, and describe what you see there.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

"

&#This looks good. Let me know if you have any questions. &#

#$&*

course Phy 121

002. Describing Graphs*********************************************

Question: `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Note that you should do these graphs on paper without using a calculator. None of the arithmetic involved here should require a calculator, and you should not require the graphing capabilities of your calculator to answer these questions.

Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points.

Now make a table for and graph the function y = 3x - 4.

Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Linear, running steeply from the lower left quadrant through the lower right quadrant and continuing on in the right upper quadrant

Intercepts x axis between 0 and 1 or at (4/3, 0). Intercepts y axis at (0, -4).

confidence rating #$&* 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`aThe graph goes through the x axis when y = 0 and through the y axis when x = 0.

The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3.

The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4).

Your graph should confirm this.

*********************************************

Question: `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The line is linear, so the slope or steepness stays the same throughout the graph

confidence rating #$&* 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`aThe graph forms a straight line with no change in steepness.

STUDENT COMMENT

Ok, I may not understand what exactly it meant by steepness, I was thinking since it was

increasing it would also be getting steeper?????

INSTRUCTOR RESPONSE

A graph can increase while getting steeper and steeper; or it can increase while getting less and less steep. Or it can increase with no change in steepness.

Analogies:

When you walk up a hill, typically as you approach the top the slope starts to level off--it gets less steep.

When you go up a ramp the steepness stays the same until you get to the end of the ramp.

When you start climbing a hill, typically it gets steeper for awhile, the stays at about a constant slope, then gets less steep toward the top.

*********************************************

Question: `q003. What is the slope of the graph of the preceding two exercises (the function is y = 3x - 4;slope is rise / run between two points of the graph)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Rise/run =

3/1 = 3

confidence rating #$&* 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`aBetween any two points of the graph rise / run = 3.

For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3.

Note that 3 is the coefficient of x in y = 3x - 4.

Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.

EXPANDED EXPLANATION

Any student who has completed Algebra I and Algebra II should be familiar with slope calculations. Most students are. However a number of students appear to be very fuzzy on the concept, and I suspect that not all prerequisite courses cover this concept adequately (though I am confident that it's done well at VHCC). Also a number of students haven't taken a math course in awhile, and might simply be a bit rusty with this idea. In any case the following expanded explanation might be helpful to some students:

Slope = rise / run.

The rise between two graph points is the change in the y coordinate. The run is the change in the x coordinate.

Our function is y = 3 x - 4.

When x = 2, we substitute 2 for x to get y = 3 * 2 - 4, which is equal to 2.

When x = 8, we substitute 8 for x to get y = 3 * 8 - 4, which is equal to 20.

The graph therefore contains the points (2, 2) and (8, 20).

You should have made a graph showing these points. If not you should do so now.

As you go from point to point your y coordinate goes from 2 to 20. So the 'rise' between the points is 20 - 2 = 18.

Your x coordinate goes from 2 to 8. So the 'run' between the points is 8 - 2 = 6.

The slope is rise / run = 18 / 6 = 3.

The numbers 2 and 8, which were used for the x values, were chosen arbitrarily. Any other two x values would have given you

different coordinates, likely with different rise and run. However whatever two x values you use, you will get the same slope. The slope of this graph is constant, and is equal to 3.

STUDENT QUESTION

Am I not allowed to utilize my calculus tools, yet?

Couldn't I have just taken the derivative for the function, y = 3x -4 to obtain 3 as the slope?

However, I do know how to do both ways. Which is the more preferred method?

INSTRUCTOR RESPONSE

This exercise develops a language for describing some aspects of graphs, and does not assume calculus tools.

Of course it's fine to use the calculus tools if you have them, as long as you understand the problem at the more basic level as well.

Unfortunately, not every student who has had a calculus course would know how to apply those tools to this situation (for example, I've had students from other institutions who have made A's in Applied Calculus courses from other (not particularly reputable) institutions, who didn't understand the concept of a slope).

*********************************************

Question: `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3.

Would you say that the graph is increasing or decreasing?

Does the steepness of the graph change and if so, how?

Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

X Y

-3 9

-2 4

-1 1

0 0

1 1

2 4

3 9

The graph decreases and then increases again with the same shape on either side of the x-axis. This graph makes a U shape with the apex at (0,0). The graph changes at the same exponential on either side of the x-axis.

confidence rating #$&* 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`aGraph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right.

The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate

STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate?

INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1?

In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&.

STUDENT QUESTION: I am a little hazy on what the steepness is

INSTRUCTOR RESPONSE: The hill analogy I used above might be helpful.

Formally, steepness could be defined as the magnitude of the slope, i.e., the absolute value of the slope.

Two graphs with respective slopes 4 and -4 would be equally steep; both would have slope of magnitude 4. Both of these graphs would be steeper than, say a graph with slope 3 or -3.

NOTE FOR STUDENT WITH CALCULUS BACKGROUND (students who haven't had calculus should ignore this; this explanation is optional even for students who have had calculus)

In terms of the calculus, the derivative function is easily seen to be y ' = 2 x, which is positive and increasing, and which therefore implies an increasing slope.

Since in this case the slope is positive, which implies that the function is increasing, the increasing slope therefore implies that the value of the function is increasing at an increasing rate.

Another terminology which is standard in calculus: If the slope is increasing then the shape of the graph is concave upward. So we could describe this graph as increasing and concave upward.

This could also be explained in terms of the second derivative, y '' = 2, which is positive everywhere. The positive second derivative implies that the graph is concave up.

*********************************************

Question: `q005. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = -3 and x = 0.

Would you say that the graph is increasing or decreasing?

Does the steepness of the graph change and if so, how?

Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

X Y

-3 9

-2 4

-1 1

0 0

The graph decreases until it reaches (0,0). The steepness changes….. from steeper, to less steep, to even less steep, or exponentially less steep…. If that makes sense. The graph is decreasing at a decreasing rate.

confidence rating #$&* 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`aFrom left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing.

Thus the graph is decreasing, but more and more slowly. We therefore say that the graph on this interval is decreasing at a decreasing rate.

NOTE FOR STUDENT WITH CALCULUS BACKGROUND (students who haven't had calculus should ignore this; this explanation is optional even for students who have had calculus)

In terms of the calculus, the derivative function is easily seen to be y ' = 2 x, which is positive and increasing, and which therefore implies an increasing slope.

Since in this case the slope is negative, which implies that the function is decreasing, the increasing slope therefore implies that the rate of decrease is decreasing. The value of the function is therefore decreasing at a decreasing rate.

Another terminology which is standard in calculus: If the slope is increasing then the shape of the graph is concave upward. So we could describe this graph as decreasing and concave upward.

This could also be explained in terms of the second derivative, y '' = 2, which is positive everywhere. The positive second derivative implies that the graph is concave up.

*********************************************

Question: `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3.

Would you say that the graph is increasing or decreasing?

Does the steepness of the graph change and if so, how?

Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

X Y

0 0

1 1

2 1.4

3 1.7

The graph increases starting at (0,0) and then rising slowly above the x-axis at it runs along the y-axis in the right upper quadrant

The steepness does change. It goes from steeper to less steep and eventually almost looking like a flat line above the y-axis

The graph is increasing at a decreasing rate

confidence rating #$&* 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`aIf you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing.

The graph would be increasing at a decreasing rate.

If the graph represents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing.

If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as described take another look at your plot and make a note in your response indicating any difficulties.

STUDENT QUESTION: I am still unsure why the steepness is decreasing, I see why going

from right to left, but the graph looks linear?

INSTRUCTOR RESPONSE: The y value increases, but it changes by less and less for every succeeding x value. So the graph is increasing, but by less and less with each step. It's increasing but at a decreasing rate.

The graph does not look linear. If it does, then it's probably because your x and/or y axis is not scaled in equal increments.

NOTE FOR CALCULUS-PREPARED STUDENTS (students who haven't had calculus should ignore this; this explanation is optional even for students who have had calculus)

In terms of the calculus, the derivative function is easily seen to be y ' = 1 / (2 sqrt(x)), which is positive but decreasing, and which therefore implies a decreasing slope.

Since in this case the slope is positive, which implies that the function is increasing, the decreasing slope therefore implies that the rate of increase is decreasing. The value of the function is therefore increasing at a decreasing rate.

Another terminology which is standard in calculus: If the slope is increasing then the shape of the graph is concave upward. So we could describe this graph as increasing and concave downward.

This could also be explained in terms of the second derivative, y '' = -1 / (4 x^(3/2)), which is negative on this interval. The negative second derivative implies that the graph is concave down.

*********************************************

Question: `q007. Make a table of y vs. x for y = 5 * 2^(-x). Graph y = 5 * 2^(-x) between x = 0 and x = 3.

Would you say that the graph is increasing or decreasing?

Does the steepness of the graph change and if so, how?

Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

X Y

0 5

1 2.5

2 1.25

3 0.625

From x = 0 to x = 3, the graph is decreasing.

The steepness changes, it becomes less stepp.

The steepness decreases at a decreasing rate

confidence rating #$&* 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a** From basic algebra recall that a^(-b) = 1 / (a^b).

So, for example:

2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4.

5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc.

The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time.

The graph is therefore decreasing at a decreasing rate. **

STUDENT QUESTION

I don’t understand how the graph decreases at a decreasing rate because it decreases by half every time. The ½ is constant.

INSTRUCTOR RESPONSE

The values decrease by a factor of 1/2 every time. That means each number would be multiplied by 1/2 to get the next.

As a result the numbers we are halving keep decreasing.

Half of 5 is 2.5; half of 2.5 is 1.25; half of 1.25 is .625. The decreases from one number to the next are respectively 2.5, 1.25 and .625.

If the y values 5, 2.5, 1.25, .625 are placed at equal x intervals, it should be clear that the graph is decreasing at a decreasing rate.

*********************************************

Question: `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster.

If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing?

Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The car travels more distance with each second, while each second remains the same time interval. So the graph increases.

The graph would increase at an increasing rate

confidence rating #$&* 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a** The speed of the car increases so it goes further each second. On a graph of distance vs. clock time there would be a greater change in distance with each second, which would cause a greater slope with each subsequent second. The graph would therefore be increasing at an increasing rate. **

STUDENT COMMENT

I don’t fully understand a distance vs. time graph.

INSTRUCTOR RESPONSE

If y represents the distance from you to the car and t represents the time in seconds since the car started out, then the graph of y vs. t is a graph of distance vs. clock time.

The car is speeding up, so in any series of equal time intervals it moves further with each new interval.

The distance it moves on an interval is represented by the difference between the y coordinates, so if it move further during an interval the 'rise' of the graph on that interval will be greater. If the intervals are equally spaced along the t axis, the result is an increasing graph with increasing slope.

This is best understood by sketching the graph according to this description.

STUDENT QUESTION

I understand the clock time but could you give me some examples of numbers to sketch a graph. I am drawing a

blank to how to make myself understand.?????

INSTRUCTOR RESPONSE

If the car's velocity for the first second averages 1 ft / sec, then in subsequent second 3 ft / sec, then 5 ft / sec, then 7

ft / sec, it will move 1 foot during the first second, 3 feet during the next, 5 feet during the next and 7 feet during the

next.

A graph of velocity vs. clock time would be a straight line, since the velocity increases by the same amount every second.

However the positions of the car, as measured from the starting point, would be

position 1 foot after 1 second

position 4 feet after 2 seconds (the position changes by 3 feet, started this second at 1 ft, so the car ends up with

position 4 feet)

position 9 feet after 3 seconds (the position changes by 5 feet, started this second at 4 ft, so the car ends up with

position 9 feet)

position 16 feet after 4 seconds (the position changes by 7 feet, started this second at 9 ft, so the car ends up with

position 16 feet)

So the graph of position vs. clock time has positions 0, 1, 4, 9 and 16 feet after 0, 1, 2, 3 and 4 seconds, respectively.

The position vs. clock time graph is therefore increasing at an increasing rate.

Let me know if this doesn't answer your question.

STUDENT QUESTION

I still don’t totally understand why it would necessarily be increasing at an increasing rate. Couldn’t it be a decreasing

or even a standard rate as I mentioned above?

INSTRUCTOR RESPONSE

If the car speeds up then its distance from its starting position increases at an increasing rate.

Its speed might be increasing at an increasing, constant or decreasing rate, but not its position.

"

&#Good responses. Let me know if you have questions. &#

#$&*

course Phy 121

006. Physics*********************************************

Question: `q001. There are two parts to this problem. Reason them out using common sense.

If the speed of an automobile changes by 2 mph every second, then how long will it take the speedometer to move from the 20 mph mark to the 30 mph mark?

Given the same rate of change of speed, if the speedometer initially reads 10 mph, what will it read 7 seconds later?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

5 seconds to get from 20mph to 30 mph (definitely not your average v-8 )

7 seconds after reading 10 mph, the car should go 24 mph

confidence rating #$&* 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds. It then repeats the process, this time passing the milepost at a speed of 20 mph.

Will the vehicle require more or less than 10 seconds to reach the lamppost?

Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The second time, the vehicle will need less than 10 seconds to get to the second lamppost. Since the initial rate of speed was greater….

The speed at the second lamppost will probably be less than 10 mph greater than before because the time between the lampposts is less, and 2mph/sec gain would lead to less gain.

confidence rating #$&* 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q003. The following example shows how we can measure the rate at which an automobile speeds up: If an automobile speeds up from 30 mph to 50 mph as the second hand of a watch moves from the 12-second position to the 16-second position, and its speed changes by 20 mph in 4 seconds. This gives us an average rate of velocity change equal to 20 mph / 4 seconds = 5 mph / second.

We wish to compare the rates at which two different automobiles increase their speed:

Which automobile speeds up at the greater rate, one which speeds up from 20 mph to 30 mph in five seconds or one which speeds up from 40 mph to 90 mph in 20 seconds?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The second car will be gaining at a faster rate of 2.5 miles per second as opposed to the first car which has a rate of 2 miles per second

confidence rating #$&* 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q004. If an automobile of mass 1200 kg is pulled by a net force of 1800 Newtons, then the number of Newtons per kg is 1800 / 1200 = 1.5. The rate at which an automobile speeds up is determined by the net number of Newtons per kg. Two teams pulling on ropes are competing to see which can most quickly accelerate their initially stationary automobile to 5 mph. One team exerts a net force of 3000 Newtons on a 1500 kg automobile while another exerts a net force of 5000 Newtons on a 2000 kg automobile.

Which team will win and why?

If someone pulled with a force of 500 Newtons in the opposite direction on the automobile predicted to win, would the other team then win?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Team 2 wins… with 2.5 newtons per kg of force. No, team 2 would still win with a 2.25 newtons per kg force compared to 2 newtons/kg force of team 1.

confidence rating #$&* 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q005. Both the mass and velocity of an object contribute to its effectiveness in a collision. If a 250-lb football player moving at 10 feet per second collides head-on with a 200-lb player moving at 20 feet per second in the opposite direction, which player do you precidt will be moving backward immediately after the collision, and why?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The 250 lb player will move backward. The two players would be venly matched at those speeds if the one running at 20 ft/sec weighed half of 250 or 125…. Since the speed of the smaller polayer is twice that of the first player. But, since the second player weighs more than 125, and is traveling at twice the first players speed, mass and velocity deems the first player to travel backwards

confidence rating #$&* 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q006. Two climbers eat Cheerios for breakfast and then climb up a steep mountain as far as they can until they use up all their energy from the meal. All other things being equal, who should be able to climb further up the mountain, the 200-lb climber who has eaten 12 ounces of Cheerios or the 150-lb climber who has eaten 10 ounces of Cheerios?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

12oz / 200 = 0.06

10 oz/ 150 = 0.066

The 150 lb climber gets further… since he got more energy from his portion of cheerios

confidence rating #$&* 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q007. Two automobiles are traveling up a long hill with an steepness that doesn't change until the top, which is very far away, is reached. One automobile is moving twice as fast as the other. At the instant the faster automobile overtakes the slower their drivers both take them out of gear and they coast until they stop.

Which automobile will take longer to come to a stop? Will that automobile require about twice as long to stop, more than twice as long or less than twice as long?

Which automobile will have the greater average coasting velocity? Will its average coasting velocity by twice as great as the other, more than twice as great or less than twice as great?

Will the distance traveled by the faster automobile be equal to that of the slower, twice that of the slower or more than twice that of the slower?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Lets say the faster vehicle is going 20 mph, and the slower one is going 10 mph. When in neutral, the 20 mph car will take longer to coast to a stop and travel a further distance because it has more velocity. The 20 mph car should take twice as long to stop, provided all other variables are constant.

confidence rating #$&* 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q008. When a 100 lb person hangs from a certain bungee cord, the cord stretches by 5 feet beyond its initial unstretched length. When a person weighing 150 lbs hangs from the same cord, the cord is stretched by 9 feet beyond its initial unstretched length. When a person weighing 200 lbs hangs from the same cord, the cord is stretched by 12 feet beyond its initial unstretched length.

Based on these figures, would you expect that a person of weight 125 lbs would stretch the cord more or less than 7 feet beyond its initial unstretched length?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The cord stretcher 4 ft with the first 50 lb difference and 3 ft with the second 50 lb increase. Average is 3.5 feet. 25 lbs added to 100lbs, or halfway between 100 and 150 lbs, would lead to 5 feet + ( 3 to 4 feet) beyond the unstretched cord. Or around 8 feet.

confidence rating #$&* 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q009. When given a push of 10 pounds, with the push maintained through a distance of 4 feet, a certain ice skater can coast without further effort across level ice for a distance of 30 feet. When given a push of 20 pounds (double the previous push) through the same distance, the skater will be able to coast twice as far, a distance of 60 feet. When given a push of 10 pounds for a distance of 8 feet (twice the previous distance) the skater will again coast a distance of 60 feet.

The same skater is now accelerated by a sort of a slingshot consisting of a bungee-type cord slung between two posts in the ice. The cord, as one might expect, exerts greater and greater force as it is pulled back further and further. Assume that the force increases in direct proportion to pullback (ie.g., twice the pullback implies twice the force).

When the skater is pulled back 4 feet and released, she travels 20 feet. When she is pulled back 8 feet and released, will she be expected to travel twice as far, more than twice as far or less than twice as far as when she was pulled back 4 feet?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

When pushed, twice the force equaled twice the distance. So when she is pulled back 8 feet, she should, technically, travel 40 feet, or twice the distance from being pulled back 4 feet.

confidence rating #$&* 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q010. Two identical light bulbs are placed at the centers of large and identically frosted glass spheres, one of diameter 1 foot and the other of diameter 2 feet.

To a moth seeking light from half a mile away, unable to distinguish the difference in size between the spheres, will the larger sphere appear brighter, dimmer or of the same brightness as the first?

To a small moth walking on the surface of the spheres, able to detect from there only the light coming from 1 square inch of the sphere, will the second sphere appear to have the same brightness as the first, twice the brightness of the first, half the brightness of the first, more than twice the brightness of the first, or less than half the brightness of the first?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

To the moth half a mile away, the first light bulb, closer to the frosted glass will seem brighter.

To the moth walking on the surface, the bulb closer to the glass, will seem brighter ( not to mention hotter). Should seem twice as bright as the first

confidence rating #$&* 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q011. The water in a small container is frozen in a freezer until its temperature reaches -20 Celsius. The container is then placed in a microwave oven, which proceeds to deliver energy at a constant rate of 600 Joules per second. After 10 seconds the ice is still solid and its temperature is -1 Celsius. After another 10 seconds a little bit of the cube is melted and the temperature is 0 Celsius. After another minute most of the ice is melted but there is still a good bit of ice left, and the ice and water combination is still at 0 Celsius. After another minute all the ice is melted and the temperature of the water has risen to 40 degrees Celsius.

Place the following in order, from the one requiring the least energy to the one requiring the most:

Increasing the temperature of the ice by 20 degrees to reach its melting point.

Melting the ice at its melting point.

Increasing the temperature of the water by 20 degrees after all the ice melted.

At what temperature does it appear ice melts, and what is the evidence for your conclusion?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Least energy = raising the temp 20 degrees to melting point

More energy = increasing the temp mof the water after the ice melted

Most energy = melting the ice at its melting point

confidence rating #$&* 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q012. Suppose you are in the center of a long, narrow swimming pool (e.g., a lap pool). Two friends with kickboards are using them to push waves in your direction. Their pushes are synchronized, and the crests of the waves are six feet apart as they travel toward you, with a 'valley' between each pair of crests. Since your friends are at equal distances from you the crests from both directions always reach you at the same instant, so every time the crests reach you the waves combine to create a larger crest. Similarly when the valleys meet you experience a larger valley, and as a result you bob up and down further than you would if just one person was pushing waves at you.

Now if you move a bit closer to one end of the pool the peak from that end will reach you a bit earlier, and the peak from the other end will reach you a little later. So the peaks won't quite be reaching you simultaneously, nor will the valleys, and you won't bob up and down as much. If you move far enough, in fact, the peak from one end will reach you at the same time as the valley from the other end and the peak will 'fll in' the valley, with the result that you won't bob up and down very much.

If the peaks of the approaching waves are each 6 inches high, how far would you expect to bob up and down when you are at the center point?

How far would you have to move toward one end or the other in order for peaks to meet valleys, placing you in relatively calm water?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

If the peaks are 6” high, the farthest you could bob up or down is 6”. I’m not sure how far you would have to move to get into calm water, but I’m assuming to the far end away from the kicking friends, since waves would still reach that end of the pool.

confidence rating #$&* 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`aIf the two 6-inch peaks meet and reinforce one another completely, the height of the 'combined' peak will be 6 in + 6 in = 12 in.

If for example you move 3 ft closer to one end you move 3 ft further from the other and peaks, which are 6 ft apart, will still be meeting peaks. [ Think of it this way: If you move 3 ft closer to one end you move 3 ft further from the other. This shifts your relative position to the two waves by 6 feet (3 feet closer to the one you're moving toward, 3 feet further from the other). So if you were meeting peaks at the original position, someone at your new position would at the same time be meeting valleys, with two peaks closing in from opposite directions. A short time later the two peaks would meet at that point. ]

However if you move 1.5 ft the net 'shift' will be 3 ft and peaks will be meeting valleys so you will be in the calmest water.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating #$&*

"

&#I need to see the questions so I can be sure what your answers mean. Most of the time I can tell, but I'm dealing with information that comes in from over 1000 different files, containing a total of about 10 000 questions. While I'm familiar with the content and sequencing of the questions, having written them all, and know what I'm looking for, different students will answer these questions in different ways and I need to be able to relate your answers to the specific wording of each question. When reviewing my responses you will also need to be able to relate your answers and my comments to the specifics of the original document. So it will be important for you on future documents to insert your responses into a copy of the original document, according to instructions, without otherwise changing any of the content of the original document. This will ensure you of the best possible feedback on your work. &#

#$&*